Question
Fig. 2.1 shows a square metal sheet of non-uniform density, with a thin wooden rod fixed at the centre of one of the corners X of the sheet.
The rod has negligible mass. The mass of the metal sheet is \(2.8\,\mathrm{kg}\).
The rod is supported so that it is horizontal and the metal sheet is vertical.
(a) Define the torque of a couple. (2 marks)
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(b) When the rod is supported in such a way that it can rotate freely within its support, the sheet hangs in equilibrium with point X vertically above the rod, as shown in Fig. 2.2.
On Fig. 2.2, draw a line to indicate the range of possible positions for the centre of gravity of the metal sheet. (1 mark)
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(c) When a torque of \(3.3\,\mathrm{N\,m}\) is applied to the rod, the sheet is held in equilibrium with two of its edges horizontal, as in Fig. 2.3. Point X is at the top-left corner.
(i) Explain whether the torque applied to the rod to hold the sheet in equilibrium is clockwise or anticlockwise. (1 mark)
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(ii) Show that the centre of gravity of the sheet has a horizontal displacement of \(0.12\,\mathrm{m}\) from the rod. (1 mark)
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(d) The square metal sheet has an average density of \(3000\,\mathrm{kg\,m^{-3}}\) and a uniform thickness of \(4.0\,\mathrm{mm}\).
Show that the side length of the sheet is \(0.48\,\mathrm{m}\). (3 marks)
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(e) Use the answer in (b) and the information in (c) and (d) to determine the position of the centre of gravity of the sheet. Indicate this position on Fig. 2.3 with a point labelled Y. (2 marks)
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 4.2: Equilibrium of forces — part (c)(ii)
• 4.3: Density and pressure — part (d)
▶️ Answer/Explanation
(a) Torque of a couple [2 marks]
The torque of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces.
\(\tau=Fd\)
Answer: \( \boxed{\text{torque of a couple}=\text{force}\times\text{perpendicular distance between the forces}} \)
(b) Possible positions of the centre of gravity [1 mark]
Since the sheet is hanging freely in equilibrium, its centre of gravity must lie vertically below the point of support.
The support is at the centre of the rod. Therefore, the possible centre of gravity lies on a straight vertical line drawn from the centre of the rod towards the bottom corner of the sheet.
Answer: \( \boxed{\text{A straight vertical line from the centre of the rod to the bottom corner}} \)
(c)(i) Direction of applied torque [1 mark]
The centre of gravity is to the right of the rod.
The weight of the sheet therefore produces a clockwise moment about the rod.
To keep the sheet in equilibrium, the applied torque must act in the opposite direction.
Answer: \( \boxed{\text{anticlockwise}} \)
(c)(ii) Horizontal displacement of the centre of gravity [1 mark]
For rotational equilibrium, the clockwise moment due to the weight must equal the applied anticlockwise torque.
\(\tau=mgd\)
\(3.3=2.8\times9.81\times d\)
\(d=\dfrac{3.3}{2.8\times9.81}\)
\(d=0.12\,\mathrm{m}\)
Answer: \( \boxed{0.12\,\mathrm{m}} \)
(d) Side length of the sheet [3 marks]
Using the density equation,
\(\rho=\dfrac{m}{V}\)
The thickness is \(4.0\,\mathrm{mm}=4.0\times10^{-3}\,\mathrm{m}\).
If the side length is \(L\), the volume is
\(V=(4.0\times10^{-3})L^2\)
Therefore,
\(3000=\dfrac{2.8}{(4.0\times10^{-3})L^2}\)
\(L=\sqrt{\dfrac{2.8}{3000\times0.0040}}\)
\(L=0.48\,\mathrm{m}\)
Answer: \( \boxed{0.48\,\mathrm{m}} \)
(e) Position of the centre of gravity [2 marks]
From part (c)(ii), the centre of gravity is \(0.12\,\mathrm{m}\) horizontally to the right of the rod.
From part (b), the centre of gravity must also lie on the vertical line through the centre of the rod when the sheet is freely supported.
Using the side length \(0.48\,\mathrm{m}\), the required position is found by drawing a line at \(45^\circ\) to the horizontal from the bottom-right corner to the edge of the rod. The centre of gravity also lies on a vertical line halfway between the rod and the right-hand edge.
Answer: \( \boxed{\text{Y is at the intersection of these two lines.}} \)